Question

1.Thực hiện phép tính sau một cách hợp lí:

A=\(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\) : \(\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)

2.Chứng minh rằng:Với mỗi x, y \(\in\) Q. Ta luôn có\(|x+y|\)\(\le\)\(|x|\)+\(|y|\)

Khi nào ta có đẳng thức?

Answer 1

Bài 1:

\(A=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}:\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)

\(A=\dfrac{2.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}:\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{2}{7}.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\)

\(A=\dfrac{2}{7}:\dfrac{2}{7}=1\)

Bài 2: Here

Chúc bạn học tốt!!!

Answer 2

1. Giải:

Gọi A =M : N

Ta có:M=\(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\)= \(\dfrac{2.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}\)=\(\dfrac{2}{7}\)

N=\(\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)=\(\dfrac{2.\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7.\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\)=\(\dfrac{2}{7}\)

Vậy A=M: N \(\Rightarrow\)A=\(\dfrac{2}{7}\):\(\dfrac{2}{7}\)=\(\dfrac{2}{7}\).\(\dfrac{7}{2}\)=\(\dfrac{2.7}{7.2}\)=1

2. Giải:

Với mọi x \(\in\)Q, ta luôn có \(x\) \(\le\) \(|x|\)(dấu bằng xảy ra khi x\(\ge\)0)

a)Nếu \(x+y\)\(\ge\)0 thì\(|x+y|=x+y\).

Vì \(x\le|x|,y\le|y|\)với mọi x, y\(\in\)Q nên:\(|x+y|=x+y\le|x|+|y|\)

b)Nếu x+y < 0 thì\(|x+y|=-\left(x+y\right)\)=\(-x-y\)

Mà -x\(\le\)\(|x|\), -y\(\le\)\(|y|\) nên: \(|x+y|\)= -x-y\(\le\)\(|x|+|y|\)

Vậy với mọi x, y\(\in\)Q ta đều có:\(|x+y|\le|x|+|y|\). Dấu bằng xảy ra khi x, y cùng dấu hoặc ít nhất có một số bằng 0.

Answer 3

2)

\(\forall x;y\in Q\)

Ta có:

\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)

\(y\le\left|y\right|và-y\le\left|y\right|\)

\(\Rightarrow x+y\le\left|x\right|+\left|y\right|\) ;\(-x-y\le\left|x\right|+\left|y\right|\)

Hay \(\left(x+y\right)\ge-\left(\left|x\right|+\left|y\right|\right)\)

Do đó\(\left|x\right|+\left|y\right|\ge\left(x+y\right)\ge-\left(\left|x\right|+\left|y\right|\right)\)

Vậy \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)

Dấu "=" xảy ra \(\Leftrightarrow xy=0\)